Class 6 Maths Chapter 8 Decimals
Question 1. 0.7499 lies between
- 0.7 and 0.74
- 0.75 and 0.79
- 0.749 and 0.75
- 0.74992 and 0.75
Solution: (3) On observing the given options, we notice that 0.7499 lies between 0.749 and 0.75.
[As 0.7499 is the successor of 0.749 and the predecessor of 0.75]
Question 2. 0.023 lies between
- 0.2 and 0.3
- 0.02 and 0.03
- 0.03 and 0.029
- 0.026 and 0.024
Solution: (2): On observing the given options, we notice that 0.023 lies between 0.02 and 0.03.
Question 3. 0.07 + 0.008 is equal to
- 0.15
- 0.015
- 0.078
- 0.78
Solution: (3): We have to find the value of 0.07 + 0.008
Read and Learn More Class 6 Maths Exemplar Solutions
First converting the given decimals into like decimals, we get 0.070 and 0.008
By writing these decimals in column form and adding them, we get
⇒ \(\begin{array}{r}
0.070 \\
+0.008 \\
\hline 0.078 \\
\hline
\end{array}\)
Hence, 0.07 + 0.008 is 0.078
Question 4. Which of the following decimals is the greatest?
- 0.182
- 0.0925
- 0.29
- 0.038
Solution: (3): First converting the given decimals into like decimals, we get 0.1820, 0.0925, 0.2900 and 0.0380.
Thus, on comparing the above, 0.2900 is the greatest among the given decimals
Question 5. Which of the following decimals is the smallest?
- 0.27
- 1.5
- 0.082
- 0.103
Solution: (3): First converting the given decimals into like decimals, we get 0.270, 1.500, 0.082, and 0.103.
Thus, on comparing the above, 0.082 is the smallest among the given decimals.
Question 6. 1 3.572 correct to the tenth place is
- 10
- 13.57
- 14.5
- 13.6
Solution: (4): When 13.572 is estimated correctly to the tenth place, we get 13.6
Question 7. 15.8 -6.73 is equal to
- 8.07
- 9.07
- 9.13
- 9.25
Solution: (B): We have given, 15.8- 6.73. Firstly converting the given decimals into like fractions, then writing the decimals in column form and subtracting, we get
∴ \(\begin{array}{r}
15.80 \\
-6.73 \\
\hline 9.07 \\
\hline
\end{array}\)
Question 8. The decimal 0.238 is equal to the fraction
- \(\frac{119}{500}\)
- \(\frac{238}{25}\)
- \(\frac{119}{25}\)
- \(\frac{119}{50}\)
Solution:
(A): We have given 0.238, which can be written in the form of
⇒ \(\frac{238}{1000}=\frac{238 \div 2}{1000 \div 2}\left[\begin{array}{l}
\text { On dividing the numerator } \\
\text { and denominator by } 2
\end{array}\right]\)
∴ \(=\frac{119}{500}\)
Question 9. \(9+\frac{2}{10}+\frac{6}{100}\) is equal to the decimal number______.
Solution:
9.26: The L.C.M. of 10 and 100 is 100.
⇒ \(9+\frac{2}{10}+\frac{6}{100}\)
⇒ \(=9 \times \frac{100}{100}+\frac{2}{10} \times \frac{10}{10}+\frac{6}{100}\)
⇒ \(\frac{900+20+6}{100}=\frac{926}{100}=9.26\)
Question 10. Decimal 1 6.25 is equal to the fraction______.
Solution:
⇒ \(\frac{65}{4}:\) We have
⇒ \(16.25=\frac{1625}{100}=\frac{1625 \div 25}{100 \div 25}\)
⇒ \(\frac{65}{4} \text { or } 16 \frac{1}{4}\)
Question 11. Fraction \(\frac{7}{25}\) is equal to the decimal number _______.
Solution:
0.28: We have, \(\frac{7}{25}=\frac{7}{25} \times \frac{4}{4}=\frac{28}{100}=0.28\)
Question 12. 4.55 + 9.73 = _______.
Solution: 14.28 : 4.55 + 9.73 = 14.28
Question 13. 8.76-2.68 = _______.
Solution: 6.08: 8.76- 2.68 = 6.08
Question 14. The value of 50 coins of 50 paisa = ₹ _______.
Solution:
25 : The value of 50 coins of 50 paisa = 50 x 50 paisa
= 2500 paisa \(=₹ \frac{2500}{100}\)
= ₹ 25
Question 15. 3 Hundredths + 3 tenths = _______.
Solution: 0.33 : 3 hundredths + 3 tenths = 0.33
Question 16. The place value of a digit at the tenth place is 10 times the same digit at the one’s place.
Solution: False
Since the place value of a digit at the tenth place is \(\frac{1}{10}\) times the same digit at the one’s place
∴ \(\frac{1}{10}=10x\)
Question 17. The place value of a digit at the hundredth place is \(\frac{1}{10}\) times the same digit at the tenth place.
Solution: True
Question 18. The decimal 3.725 is equal to 3.72 correct to two decimal places.
Solution: False
Since the digit at the thousandth place of 3.725 is 25
3.725 can be written as 3.73 (after rounding off at hundredths place)
Question 19. In the decimal form, fraction \(\frac{25}{8}=3.125\)
Solution: True
⇒ \(\frac{25}{8}=\frac{25 \times 125}{8 \times 125}\)
On multiplying the numerator and denominator by 125
∴ \(\frac{3125}{1000}=3.125\)
Question 20.The decimal 23.2 = \(23 \frac{2}{5}\)
Solution: False
We have \(23.2=\frac{232}{10}=\frac{116}{5}\)
⇒ \(23 \frac{1}{5}\)
Question 21. 3.03 + 0.016 = 3.019
Solution: False
Firstly convert 3.03 and 0.016 into like fractions, writing the decimals in column form and finally by adding we get,
⇒ \(\begin{array}{r}
3.030 \\
+0.016 \\
\hline 3.046
\end{array}\)
Question 22. 42.28-3.19 = 39.09
Solution: True
⇒ \(\begin{array}{r}
42.28 \\
-3.19 \\
\hline 39.09 \\
\hline
\end{array}\)
Question 23. 19.25 <19.053
Solution: False
Since, the digit at a tenth place of 19.25 is 2 and the digit at a tenth place of 19.053 is 0, where 2>0.
19.25 > 19.053
Question 24. 13.730 = 13.73
Solution: True
Since, after converting the given decimals into like decimals we get, 13.730 = 13.73.
Question 25. 3.25…3.4
Solution:
<: Converting the given decimals into like decimals, they become 3.25 and 3.40. The whole number part of these is the same. On comparing, we get their tenth digits 2 < 4
3.25 < 3.4
Question 26. 6.25…\( \frac{25}{4}\)
Solution:
=: \(6.25=\frac{625}{100}=\frac{625 \div 25}{100 \div 25}=\frac{25}{4}\)
∴ \(6.25\frac{25}{4}\)
Question 27. Arrange 12.142, 12.124, 12.104, 12.401 and 12.214 in ascending order.
Solution:
Given
12.142, 12.124, 12.104, 12.401 and 12.214
The digits are already given in the form of like decimals
Clearly,
12.104 < 12.124 < 12.142 < 12.214 < 12.401
Question 28. Write the largest four-digit decimal number less than 1 using the digits 1, 5, 3 and 8 once.
Solution:
The required number is 0.8531, which is the largest four-digit decimal number less than 1.
Question 29. Using the digits 2, 4, 5 and 3 once, write the smallest four-digit decimal number.
Solution:
The required number is 0.2345, which is the smallest four-digit decimal number
Question 30. Express \(\frac{11}{20}\) as a decimal.
Solution:
Given
latex]\frac{11}{20}[/latex]
We have, \(\frac{11}{20}=\frac{11 \times 5}{20 \times 5}=\frac{55}{100}=0.55\)
Question 31. Express \(3 \frac{2}{5}\) as a decimal.
Solution:
Given
\(3 \frac{2}{5}\)We have \(3 \frac{2}{5}=\frac{(3 \times 5)+2}{5}=\frac{15+2}{5}=\frac{17}{5}\)
Now, \(\frac{17}{5}=\frac{17 \times 2}{5 \times 2}=\frac{34}{10}=3.4\)
Question 32. Express 0.041 as a fraction.
Solution:
Given 0.041
We have, \(0.041=\frac{41}{1000}\)
Question 33. Express 6.03 as a mixed fraction.
Solution:
Given 6.03
We have, \(6.03=\frac{603}{100}\)
⇒ \(6 \frac{\mathrm{3}}{100}\)
Question 34. Convert 5201 g to kg.
Solution:
Given
5201
We have ,\(5201 \mathrm{~g}=\frac{5201}{1000} \mathrm{~kg}\)
[1kg = 1000 g]
= 5.201 kg
Question 35. Convert the 2009 paise to rupees and express the result as a mixed fraction.
Solution:
We have 2009 paise \(=₹ \frac{2009}{100}=₹ 20.09\)
∴ \(\text { or } ₹ 20 \frac{9}{100}\)
36. Convert 1 537 cm to m and express the result as an improper fraction.
Solution:
We have, 1537 cm \(=\frac{1537}{100} \mathrm{~m}\)
[1 m = 100 cm]
= 15.37 m
Question 37. Convert 2435 m to km and express the result as a mixed fraction
Solution:
Given 2435
we have, 2453 m \(=\frac{2435}{1000} \mathrm{~km}\)
[1 km = 1000 m]
= 2.435 km
Firstly, convert the fraction \(\frac{2435}{1000}\) into the simplest form, for dividing the numerator and denominator by 5, we get
⇒ \(\frac{2435 \div 5}{1000 \div 5} \mathrm{~km}=\frac{487}{200} \mathrm{~km}\)
∴ \(2 \frac{87}{200} \mathrm{~km}\)
Question 38. Round off 20.83 to the nearest tenths.
Solution: The estimated value of 20.83 to the nearest tenths is 20.8
Question 39. Round off 75.1 95 to the nearest hundredths.
Solution: The estimated value of 75.195 to the nearest hundredths is 75.20
Question 40. Round off 27.981 to the nearest tenths.
Solution: The estimated value of 27.981 to the nearest tenths is 28.0
Question 41. What should be added to 25.5 to get 50?
Solution: According to question, 25.5 +? = 50
⇒ \(\begin{array}{r}
50.0 \\
-25.5 \\
\hline 24.5 \\
\hline
\end{array}\)
? = 50 – 25.5
= 24.5
Thus, the required number is 24.5
Question 42. Alok purchased 1 kg 200 g potatoes, 250 g dhania, 5 kg 300 g onion, 500 g palak and 2 kg 600 g tomatoes. Find the total weight of his purchases in kilograms.
Solution:
Given
Alok purchased 1 kg 200 g potatoes, 250 g dhania, 5 kg 300 g onion, 500 g palak and 2 kg 600 g tomatoes.
Alok purchased,
Potatoes =1 kg 200 g = 1.200 kg
Dhania = 250 g = 0.250 kg
Onion = 5 kg 300 g = 5.300 kg
Palak = 500 g = 0.500 kg
Tomatoes = 2 kg 600 g = 2.600 kg
The total weight of the above purchases
⇒ \(\begin{array}{r}
1.200 \mathrm{~kg} \\
0.250 \mathrm{~kg} \\
5.300 \mathrm{~kg} \\
0.500 \mathrm{~kg} \\
+2.600 \mathrm{~kg} \\
\hline \underline{9.850 \mathrm{~kg}} \\
\hline
\end{array}\)
Question 43. Arrange in ascending order: 0.011, 1.001, 0.101, 0.110
Solution:
Given
0.011, 1.001, 0.101, 0.110
Since, all the decimals are already given in like fractions, i.e., 0.011, 1.001, 0.101, 0.110.
Arranging them in ascending order, we get 0.011, 0.101, 0.110, 1.001
Question 44. Add the following: 20.02 and 2.002
Solution:
We have, 20.02 and 2.002 To add the above decimals, we must convert them into like decimals first.
Writing 20.020 and 2.002 in a column
So,
⇒ \(\begin{array}{r}
20.020 \\
+2.002 \\
\hline 22.022 \\
\hline
\end{array}\)
which is the required sum.
Question 45. The energy content of different foods is as follows:
Which food provides the least energy and which provides the maximum? Express the least energy as a fraction of the maximum energy.
Solution:
Milk provides the least energy and rice provides the maximum energy
The required fraction = \(\frac{3.0}{5.3}=\frac{3.0 \times 10}{5.3 \times 10}\)
∴ \(=\frac{30}{53}\)
Question 46. Which one is greater? 1 metre 40 centimetres + 60 centimetres or 2.6 metres.
Solution:
1 metre 40 centimetres + 60 centimetres = 1.40 metres + 0.60 metres
[ 100 centimetres =1 metre]
= 2.00 metres
Since, 2.6 >2.00
2.6 metres is greater.